## Posts Tagged ‘minimax inequality’

### Longest shortest geodesic on a 2-sphere

October 31, 2011

This is a little note about constructing a Riemannian 2-sphere which has longer shortest geodesic than the round 2-sphere of same area.

—–  Background Story  —–

So there has been this thing called ‘mathematical conversations’ at the IAS, which involves someone present a topic that’s elementary enough to be accessible to mathematicians in all fields and yet can be expanded in different directions and lead into interesting interdisciplinary discussions.

Nancy Hingston gave one of those conversations about simple geodesics on the two-sphere one night and I was (thanks to Maria Trnkova who dragged me in) able to attend.

So she talked about some fascinating history of proving the existence of closed geodesics and later simple closed geodesics on generic Riemannian two-spheres.

Something about this talk obviously touched my ‘systolic nerve’, so when the discussion session came up I asked whether we have bounds on ‘length of longest possible shortest closed geodesic on a sphere with unit area’. The question seem to have generated some interest in the audience and resulted in a back-and-forth discussion (some of which I had no clue what they were talking about). So the conclusion was at least nobody knows such a result on top of their head and perhaps optimum is obtained by the round sphere.

—–  End of Story —-

A couple of post-docs caught me afterwards (Unfortunately I didn’t get their names down, if you happen to know who they are, tell me~) and suggested that suspending a smooth triangular region and smoothen the corners might have longer shortest geodesic than the round sphere:

The evidence being the fact that on the plane a rounded corner triangular contour has larger ‘width’ than the disc of same area. (note such thing can be made to have same width in all directions)

Well that’s pretty nice, so I went home and did a little high-school computations. The difficulty about the pillow is that the shortest geodesic isn’t necessarily the one that goes through the ‘tip’ and ‘mid-point of the base’, something else might be shorter. I have no idea how to argue that.

A suspicious short geodesic:

So I ended up going with something much simpler, namely gluing together two identical copies of the flat equilateral triangles. This can be made to a Riemannian metric by smoothing the edge and corners a little bit:

Okay, now the situation is super simple~ I want to prove that this ‘sphere’ (let’s call it $S$ from now on) has shortest geodesic longer than the round sphere ($\mathbb{S}^2$)!

Of course we suppose both $S$ and $\mathbb{S}^2$ has area 1.

Claim: The shortest geodesic on $S$ has length $\sqrt[4]{12}$ (which is length of the one through the tip and mid-point of the opposite edge.)

Proof: The shortest closed geodesic passing through the corner is the one described above, since any other such geodesics must contain two symmetric segments from the corner to the bottom edge on the two triangles, those two segments alone is longer than the one orthogonal to the edge.

That middle one has length $2h$ where

$A(\Delta) = 1/2 = h^2/\sqrt{3}$

i.e. $h = \sqrt[4]{3} / \sqrt{2}, \ \ell = 2h = \sqrt[4]{12}$

The good thing about working with flat triangles is that now I know what the closed geodesics are~

First we observe any closed geodesic not passing through the corner is a periodical billiard path in the triangular table with even period.

So let’s ‘unfold’ the triangles on the plane. Such periodic orbits correspond to connecting two corresponding points on a pair of identified parallel edges and the segment between them intersecting an even number of tiles.

W.L.O.G we assume the first point in on edge $e$. Since we are interested in orbits having shortest length, let’s take neighborhood of radius $\sqrt[4]{12} + \epsilon$ around our edge $e$: (all edges with arrows are identified copies of $e$)

There are only 6 parallel copies of $e$ in the neighborhood:

Note that no matter what point $p$ on $e$ we start with, the distance from $p$ to another copy of it on any of the six edges is EQUAL to $\sqrt[4]{12}$. (easy to see since one can slide the segments to begin and end on vertices.)

Hence we conclude there are no shorter periodic billiard paths, i.e. the shortest closed geodesic on $S$ has length $\sqrt[4]{12}$.

Note it’s curious that there are a huge amount of closed geodesics of that particular length, most of them are not even simple! However it seems that after we smoothen $S$ to a Riemannian metric, the non-simple ones all become a little longer than that simple one through the corner. I wonder if it’s possible that on a Riemannian sphere the shortest closed geodesic is a non-simple one.

Anyways, now let’s return to $\mathbb{S}^2$~ So the surface area is $1$ hence the radius is $r= \sqrt{1/4\pi} = \frac{1}{2\sqrt{\pi}}$

Any closed geodesics is a multiple of a great circle, hence the shortest geodesic has length $2 \pi r = \sqrt{\pi}$, which is just slightly shorter than $\sqrt[4]{12} \approx \sqrt{3.4}$.

Now the natural question arises: if the round sphere is not optimum, then what is the optimum?

At this point I looked into the literature a little bit, turns out this problem is quite well-studied and there is a conjecture by Christopher Croke that the optimum is exactly $\sqrt[4]{12}$. (Of course this optimum is achieved by our singular triangle metric hence after smoothing it would be $< \sqrt[4]{12}$.

There is even some recent progress made by Alex Nabutovsky and Regina Rotman from (our!) University of Toronto! See this and this. In particular, one of the things they proved was that the shortest geodesic on a unit area sphere cannot be longer than $8$, which I believe is the best known bound to date. (i.e. there is still some room to $\sqrt[4]{12}$.)

Random remark: The essential difference between this and the systolic questions is that the sphere is simply connected. So the usual starting point, namely ‘lift to universal cover’ for attacking systolic questions does not work. There is also the essential difference where, for example, the question I addressed above regarding whether the shortest geodesic is simple would not exist in systolic situation since we can always split the curve into two pieces and tighten them up, at least one would still be homotopically non-trivial. In conclusion since this question sees no topology but only the geometry of the metric, I find it interesting in its own way.

### Intergal geometry and the minimax inequality in a nutshell

February 7, 2011

The goal for most of the posts in this blog has been to take out some very simple parts of certain papers/subjects and “blow them up” to a point where anybody (myself included) can understand. Ideally the simple parts should give some inspirations and ideas towards the more general subject. This one is on the same vein. This one is based on parts of professor Guth’s minimax paper.

In an earlier post, we talked about the extremal length where one is able to bound the “largest possible minimum length” (i.e. the “maximum minimum length“) of a family of rectifiable curves under conformal transformation. When combined with the uniformization theorem in for surfaces, this becomes a powerful tool for understanding arbitrary Riemannian metrics (and for conformal classes of metrics in higher dimensions).

However, in ‘real life’ we often find what we really want to bound is, instead, the “minimum maximum length” of a family of curves, for example:

Question: Let $\mathbb{D} \subseteq \mathbb{R}^2$ be the unit disc. Given any family $\mathcal{F}$ of arcs with endpoints on $\partial \ \mathbb{D}$ and $\mathcal{F}$ foliates $\mathbb{D}$, then how short can the logest arc in $\mathcal{F}$ possibly be?

In other words, let $\mathbb{F}$ be the collection of all possible such foliations $\mathcal{F}$ as above, what is

$\displaystyle \inf_{\mathcal{F} \in \mathbb{F}} \ \sup_{A \in \mathcal{F}} \ \ell(A)$?

After playing around a little bit with those foliations, we should expect one of the fibres to be at least as long as the diameter ( i.e. no foliation has smaller maximum length leaf than foliating by straight lines ). Hence we should have

$\displaystyle \inf_{\mathcal{F} \in \mathbb{F}} \ \sup_{A \in \mathcal{F}} \ \ell(A) = 2$.

This is indeed easy to prove:

Proof: Consider the map $f: S^1 \rightarrow S^1$ where $S^1 = \partial \ \mathbb{D}$, $f$ switches the end-points of each arc in $\mathcal{F}$. It is easy to check that $f$ is a continuous, orientation reversing homeomorphism of the circle (conjugate to a reflection). Let $p, q$ be its fixed points, $L_1, L_2$ be the two arcs in $S^1$ connecting $p$ to $q$.

Let

$g: z \mapsto -z$

be the antipodal map on $S^1$.

Suppose $p \neq g(q)$ then one of $L_1, L_2$ is longer than $\pi$, say it’s $L_1$.

Then we have

$f \circ g (L_1) \subseteq L_1$.

Hence $f \circ g$ has a fixed point $m$ in $L_1$, i.e. $f(m) = -m$.

There is a fibre $A$ in $\mathcal{F}$ with endpoints $m, -m$, the fibre must have length

$\ell(A) \geq d(-m,m) = 2$.

The remaining case is trivial: if $p = g(q)$ then both $L_1$ and $L_2$ gets mapped into themselves orientation-reversingly, hence fixed points still exists.

Establishes the claim.

Instead of the disc, we may look at circles that sweep out the sphere (hence to avoid the end-point complications):

Theorem: Any one-parameter family of circles that foliates $S^2$ (except two points) must have the largest circle being longer than the equator.

This is merely applying the same argument, i.e. one of the circles needs to contain a pair of antipodal points hence must be longer than the equator.

In order for easier generalization to higher dimensions, with slight modifications, this can be formulated as:

Theorem: For any $f: T^2 \rightarrow S^2$ having non-zero degree, there is $\theta \in S^1$ where $\ell(f(S^1 \times \{ \theta \})$ is larger than the equator.

Hence in higher dimensions we can try to prove the same statement for largest image of a lever $k$-sphere under $f: S^k \times S^{n-k} \rightarrow S^n$. However before we do that I would like to highlight some intergal geometry machineries that are new to me but seemingly constantly used in proving those kinds of estimates. We shall get some idea of the method by showing:

Theorem: Let $\mathbb{R}P^n$ be equipped with the round metric. $p^k \subseteq \mathbb{R}P^n$ be a ‘flat’ $k$-dimensional plane. Then any $k$-chain $z^k \subseteq \mathbb{R}P^n$ in the same $k$ dimensional homology class as $p^k$ must have volume at least as large as $p^k$.

Proof: Let $Gr(\mathbb{R}P^n, n-k)$ be the set of all $(n-k)$-planes in $\mathbb{R}P^n$ (i.e. the Grassmannian).

There is a standard way to associate a measure $\mu$ on $Gr(\mathbb{R}P^n, n-k)$:

Let $\lambda$ be the Haar measure on $SO(n+1)$, fix some $Q \in Gr(\mathbb{R}P^n, n-k)$.

Since $SO(n+1)$ acts on $\mathbb{R}P^n$, for open set $S \subseteq Gr(\mathbb{R}P^n, n-k)$, we set

$\mu(S) = \lambda( \{ T \in SO(n+1) \ | \ T(Q) \in S \})$.

–The measure of a collection of planes is the measure of linear transformations that takes the given plane to an element of the set.

Now we are able to integrate over all $(n-k)$-planes!

For almost all $Q \in Gr(\mathbb{R}P^n, n-k)$, since $P$ is $k$-plane, we have $| Q \cap P | = 1$. ( not $1$ only when they are ‘parallel’ )

Since $[z] = [p]$ in $H_k(\mathbb{R}P^n, \mathbb{Z}_2)$, for almost all $Q$, $z$ intersects $Q$ at least as much as $P$ does. We conclude that for almost all $Q, \ | z \cap Q | \geq 1$.

Fact: There exists constant $C$ such that for any $k$-chain $\Sigma^k \in \mathbb{R}P^N$,

$\mbox{Vol}_k(\Sigma^k) = \mathbb{E}(|\Sigma \cap Q |)$.

The fact is obtained by diving the chain into fine cubes, observe that both volume and expectation are additive and translation invariant. Therefore we only need to show this for infinitesimal cubes (or balls) near $0$. We won’t work out the details here.

Hence in our case, since for almost all $Q$ we have $| z \cap Q | \geq 1$, the expectation $\mathbb{E}(|z \cap Q |) \geq 1$.

We therefore deduce

$\mbox{Vol}_k(z) = \mathbb{E}(|z \cap Q |) \geq 1$.

Establishes the theorem.

Remark: I found this intergal geometry method used here being very handy: in the old days I always try to give lower bounds on volume of stuff by intersecting it with planes and then pretend the ‘stuff’ were orthogonal to the plane, which is the worst case in terms of having small volume. An example of such bound can be found in the knot distorsion post where in order to lower bound the length we look at its intersection number with a family of parallel planes and then integrate the intersection.

This is like looking from one particular direction and record how many times did a curve go through each height, of course one would never get the exact length if we know the curve already. What if we are allowed to look from all directions?

I always wondered if we know the intersection number with not only a set of parallel planes but planes in all directions, then are there anything we can do to better bound the volume? Here I found the perfect answer to my question: by integrating over the Grassmannian, we are able to get the exact volume from how much it intersect each plane!

We get some systolic estimates as direct corollaries of the above theorem, for example:

Corollary: $\mbox{Sys}_1(\mathbb{R}P^2) = \sqrt{\pi/2}$ where $\mathbb{R}P^2$ carries the round metric with total volume $1$.

Back to our minimax problems, we state the higher dimensional version:

Wish: For any $C^1$ map $f: S^k \times S^{n-k} \rightarrow S^n$ where $S^n$ carries the standard round metric, there exists some $\theta \in S^{n-k}$ with

$\mbox{Vol}_k(f(S^k\times \{\theta\})) \geq \mbox{Vol}_k(E^k)$

where $E^k \subseteq S^n$ is the $k$-dimensional equator.

But what we have is that there is a (small) positive constant $c(n,k)$ s.t. $\mbox{deg}(f) \neq 0$ implies

$\displaystyle \sup_{\theta \in S^{n-k}} \mbox{Vol}_k(f(S^k \times \{\theta\})) \geq c(n,k) \mbox{Vol}_k(E^k)$

(shown by an inductive application of the isomperimetric inequality on $S^N$, which is obtained from applying intergal geometry methods)